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M02-21

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If \(x\) is a positive integer, is \(y^2 (x^3 - x + 1) \gt 75?\)

We need to know values of two variables to answer the question:
  1. The value of \(x\)
  2. The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.


Answer: C
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New post 11 Oct 2014, 02:55
from (2): we can also get the x=-3. Bunuel can i infer this?

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New post 11 Apr 2015, 07:06
Bunuel wrote:
Official Solution:


We need to know values of two variables to answer the question:
  1. The value of \(x\)
  2. The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.


Answer: C


I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3 - 5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = -1\). That should be the second statement of x=5 is sufficient.

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Bunuel wrote:
Official Solution:


We need to know values of two variables to answer the question:
  1. The value of \(x\)
  2. The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.


Answer: C


I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3 - 5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = -1\). That should be the second statement of x=5 is sufficient.


What if y=0? Or y=1/2? Or y=-1/2? There are infinitely many values of y, for which y^2>75/121 is not true.
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New post 14 Apr 2015, 22:50
Yes even I had the same question , but if we read the question carefully it says x is a positive integer. So, x=-3 cannot be considered in this case.

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New post 27 Feb 2016, 07:21
I think this is a high-quality question and I don't agree with the explanation. Hi,

Since y is squared in the statement, a negative integer will not matter, right? Hence, shouldn't statement 2 alone be sufficient?

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New post 29 Feb 2016, 08:10
anisharao2 wrote:
I think this is a high-quality question and I don't agree with the explanation. Hi,

Since y is squared in the statement, a negative integer will not matter, right? Hence, shouldn't statement 2 alone be sufficient?


Please read the solution more carefully. What if y=0?
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New post 22 Aug 2016, 13:09
I think this is a high-quality question and I agree with explanation.

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New post 01 Dec 2016, 08:15
I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75

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New post 01 Dec 2016, 08:43
anujjain2016 wrote:
I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75


Please read carefully: If x is a positive integer, is ...
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New post 22 Mar 2017, 03:40
Hi Bunuel,

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?

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New post 22 Mar 2017, 04:14
akshayv7 wrote:
Hi Bunuel,

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?


Not sure I understand what you mean...

The question asks whether \(y^2 (x^3 - x + 1) \gt 75\), not whether y > 0. (2) says that y > 0. From (1) the question becomes whether \(y^2 \gt 75/121\) and since from (2) we know that y > 0, then when considering the statements together we can answer that YES \(y^2 \gt 75/121\).
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New post 19 Jun 2017, 23:14
I solved statement 2 in following manner:

(x-1)^2=16
x^2-2x+1=16
x^2-2x=15
x(x-2)=15
So x=0 or x=17

Where did I go wrong?

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New post 20 Jun 2017, 00:19
disharupani wrote:
I solved statement 2 in following manner:

(x-1)^2=16
x^2-2x+1=16
x^2-2x=15
x(x-2)=15
So x=0 or x=17

Where did I go wrong?


You can equate multiples to 0, when the whole product equals 0. For example, if it were x(x - 2) = 0, then yes, x = 0 or x = 2. Notice that if x = 0, then x(x - 2) = 15, does not hold. Also, it's not clear how you got x = 17, there. Again, if x = 17, then x(x - 2) = 15, does not hold.
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New post 29 Aug 2017, 11:39
Hi Bunuel,

1 clarification needed..
in statement 2->
we get x=5
y^2 * 121>75 to be proven?
y^2 > or = to zero
even if we take y as 1/2 then y^2 =1/4
1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0
0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121
144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

Thanks,
Stuti

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New post 29 Aug 2017, 20:45
stutimittal04 wrote:
Hi Bunuel,

1 clarification needed..
in statement 2->
we get x=5
y^2 * 121>75 to be proven?
y^2 > or = to zero
even if we take y as 1/2 then y^2 =1/4
1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0
0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121
144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

Thanks,
Stuti


For some values of y, the answer is YES, for example if y = 100.
For some values of y, the answer is NO, for example if y = 0.
Two answers, hence insufficient.
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Ohok!! Thanks

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New post 24 Oct 2017, 23:08
What do you guys think about the variable approach from Math Revolution? I think that in some cases it is better to use this approach, but in most cases - not. And if you are aiming for, at least, Q50, variable approach should not be your friend

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